Looking for some help on garage sub panel

URD cannot enter either building for even one half inch. In the event of a fire the insulation is toxic.

If you want to run from panel to panel the use THW conductors, the price will be a little higher but the benefits outweigh the price.

Click to expand...

jwelectric,

I MUST be missing something..... I did contact my local town building code office, They forwarded me to the Electrical Inspector / Code Enforcement officer. I told him what i am doing, and asked if he had a recomendation on where to purchase my materials. He told me to just go to Home Depot, and get their 100 amp feed. I told him about how i was told the insulation is toxic if there was a fire, and he said that all wire insulation is toxic, and inside of conduit the intention is that my family would have more than enough time to excape........ I can only assume that there are some markings on the wire i am missing. I do know that it has URD on it, but are ther other markings that may dual rate the wire? Could the rules be different in different states?
I am not saying you are wrong. If the wire is URD only i will go another route.

I just simply like to understand as much as i can, I figure the more i know, and the more homework i do, the less mistakes i will make

........ I can only assume that there are some markings on the wire i am missing. I do know that it has URD on it, but are ther other markings that may dual rate the wire? Could the rules be different in different states?
I am not saying you are wrong. If the wire is URD only i will go another route.

I just simply like to understand as much as i can, I figure the more i know, and the more homework i do, the less mistakes i will make

Click to expand...

Most URD is dual rated- it has more than one set of letters URD, RHW or some other alphabet soup.

I would go to Home Desperate and buy 2-2-4-6 tri-rated mobile home feeder, tag it to a ninety amp breaker to feed the shop, and never look back.

Fine. Not to be pedantic: assuming all the loads are at 120v, and the panel is protected with a 50 amp breaker, then one could have a total of 100 amps of demand going.

If one had a single 240v demand of 10 amp, then the total ampacity you could deliver at 120v would be 80.

40 on each leg.

Do you have an opinion on how much ampacity this guy actually NEEDS?

Click to expand...

This is totally incorrect. The rating on the handle of a breaker is what that breaker will carry. A two pole 50 amp breaker is rated at 50 amps of 240 volts or it is rated at 50 amps per leg. They don’t add up.

Look at your dryer and range breakers and the neutral conductor that is installed with each one. If the two poles of the thirty amp breaker would add to 60 amps they why is the white conductor supplying that dryer not a #6 instead of a #10 if the two added to 60 amps?

As Dave has pointed out most URD is dual rated but a lot of times it is not.

If the conductor is not listed in table 310.104(A) of the 2011 code cycle then it can’t be used. You won’t find URD listed there because it is a utility conductor and not one that is allowed to be installed in the premises wiring by us.

Many inspectors think that if it can be bought at some supply house then it is okay to install. This if not true, it must be listed in the NEC or it can’t be used.

Is it used? Yes many times but that don’t mean that it is right. I would never install URD for anyone for any reason simply because should the insulation burn someone might die. Never and I repeat never bring URD to the inside of any building unless it is a dual rated conductor.

Edited to add;

It is fine to install the 90 amp feeder to the garage if this is what you want to do. The 90 amp breaker you install to protect the #2 conductors you are installing is just that, 90 amps. It does not add up to 180 amps of 120 volts of current. It is 90 amps period.

I am looking into the grounding rods, there are both Galvanized and Copper, I am assuming that copper is the way to go. Does 5/8" or 3/4" matter? Does the grounding rod have to be out side the structure, or can i core drill thru the sill plate, and concrete then drive the grounding rod inside the garage this way i dont have somethin to trip on, and trim around when cutting the grass.

FYI i will be borrowing a 3/4" hammer drill from my work, and making up an adaptor for driving the grounding rod.

What we do where i work is take a of pipe where the id is a little bigger than the diameter of the rod, and weld a piece of round bar to it, then if you set the hammer drill to hammer only mode it will drive the rod in super easy and fast. I looked around a little bit and we already have some made up for 5/8" and 3/4 diameters

We drive 3/4" re-bar 20 ft long in no exageration about 2 minutes. We just dig a hole about the size of a pop can, and fill it with water set the end of the re-bar in the middle of the hole and drive it down.

Where i work we build bridges, and sometimes need to drive re-bar to shore up for our crane matts.

This is totally incorrect. The rating on the handle of a breaker is what that breaker will carry. A two pole 50 amp breaker is rated at 50 amps of 240 volts or it is rated at 50 amps per leg. They don’t add up.

Look at your dryer and range breakers and the neutral conductor that is installed with each one. If the two poles of the thirty amp breaker would add to 60 amps they why is the white conductor supplying that dryer not a #6 instead of a #10 if the two added to 60 amps?

Click to expand...

Oh, jeepers.

If he has a panel that has a 50 amp breaker, then he can have 50 amp of demand at 120v on one side of the phase and 50 amp of demand at 120v on the other side of the phase.

50 plus 50 equals 100. No, he cannot have a 60 amp load on one side of the phase and only have a 40 amp demand on the other side, and have the breaker hold.

I suppose there are people out there dull enough not to get that. I am not one of them.

If he has a panel that has a 50 amp breaker, then he can have 50 amp of demand at 120v on one side of the phase and 50 amp of demand at 120v on the other side of the phase.

50 plus 50 equals 100. No, he cannot have a 60 amp load on one side of the phase and only have a 40 amp demand on the other side, and have the breaker hold.

I suppose there are people out there dull enough not to get that. I am not one of them.

Click to expand...

My friend if he has a 50 amp breaker and he has 50 amps of 120 volts on one side and he has 50 amps on the other side he would have a total of 50 amps at 240 volts and there would be no amps on the neutral.

Using your analogy of 50 amps on one side and then 50 amps on the other side there would be 100 amps on the neutral which is not true. If there was 50 amps on one side and 40 amps on the other side there would only be 10 amps on the neutral.

I suppose there are people out there dull enough not to get that and you are one of them.

I wrote nothing of the sort and I don't understand why you insist on distorting my comments.

And you can take your insults and insert them.

Click to expand...

I am not trying to distort or insult and I am sorry if you take it that way. All I am trying to do is to get you to see that a 50 amp breaker will only deliver 50 amps of current be it 240 or 120. It is not 50 plus 50 at 120 volts.

Oh, jeepers. If he has a panel that has a 50 amp breaker, then he can have 50 amp of demand at 120v on one side of the phase and 50 amp of demand at 120v on the other side of the phase.
50 plus 50 equals 100. No, he cannot have a 60 amp load on one side of the phase and only have a 40 amp demand on the other side, and have the breaker hold.
I suppose there are people out there dull enough not to get that. I am not one of them.

Click to expand...

If what you are saying that 50 plus 50 equals 100 amps then just how much do you think the neutral would be carrying?
Now if your answer is zero then there would be 50 amps on one side and 50 amps on the other side not a total of 100. All the current flow from one leg would be returning on the other leg if there was no current on the neutral.

If I used an ammeter and checked the amperage on the two legs of the feeder and found 50 amps on one leg and 40 amps on the other leg there would be 10 amps on the neutral and the entire 240 volt circuit would be drawing 50 amps not 90 amps.

There would be 40 amps of 240 volt current and 10 amps of 120 volt current even if everything on these feeders was 120 volt circuits.

If you are saying that on one side of the circuit there is 50 amps what you are saying is that there is 50 amps returning on the neutral. If you say that the other side is carrying another 50 amps then this current is returning on the neutral. If you are saying that 50 plus 50 equals 100 then this is saying that one side has 50 amps on the neutral and the other side has 50 amps on the neutral and this equals 100 amps on the neutral.

If you are saying that the two sides would balance the neutral and there would be no current on the neutral then just how is this current returning? It must be returning on one of the hot conductors.

No insults are intended just trying to get you to understand that the two donâ€™t add up.

Jw, you truly are the master of this, I always read your threads because they are always insightful, I love them. Just for the heck of it, I will tell you, for a minute there I thought I was at work with something you said. lol. You said, " I am sorry you feel that way..." I just recently, said that to a very irate person who didn't understand too, I was not trying to be rude or insulting, only helpful. Sometimes, it is what it is.

I am not trying to distort or insult and I am sorry if you take it that way.

Click to expand...

For the love of gawd, stop being so damned pedantic.

If you could uncork your illusion that you are perfect for one moment, you will see that I am just telling this guy that a "50 amp service" does not mean he can only have 50 amps of demand at 120 volts.

He has two sides of the phase and each can carry 50 amps.

So if he has two loads that each demand 45 amps at 120v he does not need a 100 amp service to accommodate them.

If you could uncork your illusion that you are perfect for one moment, you will see that I am just telling this guy that a "50 amp service" does not mean he can only have 50 amps of demand at 120 volts.

I am here to insure that any and all information given is true and safe.
A 50 amp feeder is just that, 50 amps. It is not 50 amps at 120 volts plus another 50 amps at 120 volts for a total of 100 amps at 120 volts.
If there was 50 amps at 120 volts on one side and another 50 amps at 120 volts on the other side the neutral would have 100 amps on it.
If you are going to say this is untrue then pray tell me how these amps are returning to the transformer.

Edited to add;

For the sake of this discussion letâ€™s take away the trip curve of breakers and assume that a 50 amp breaker will trip at 51 amps.

I have a panel that is supplied by 240 volts at 50 amps.

I load the panel to 50 amps of 120 volt loads and all this load is on one side of the 50 amp breaker. I now load the other side of the breaker with a wall clock and the breaker trips. The total load on that 240 volt two pole breaker will equal 50 amps plus one wall clock.

What canâ€™t be done is to install a two pole 50 amp breaker and load one side to 50 amps at 120 volts and then add another 50 amps at 120 volts to the other side of the breaker at 120 volts to where there would be a total of 100 amps of 120 volt load. It just doesnâ€™t work that way.

Take this same feeder at 50 amps and install two 100 watt lights one on each side of the breaker. The total load on that feeder will be 200 watts with or without the neutral. As long as each side of that breaker has the exact same amount of current the neutral is not needed. A simple experiment you can do is use two lights at 100 watts each and connect them to a 240 volt circuit in series without a neutral and check the voltages at each bulb. You will find that each bulb has 120 volts of drop which equals a total load on that 240 volt circuit of 1.66 amps although each bulb only pulls .83 amps each. Now install the neutral and the load on the feeders will still be 1.66 amps.

Using this same circuit that has a amperage draw of 50 amps on one leg and a 40 amps on the other leg and the neutral will have 10 amps of current flowing through it. You will have a total of 40 amps of 240 volt series circuit and 10 amps of 120 volt parallel circuit but the very second that either side reaches 51 amps the circuit will open (this is if we take away any trip curve of the breaker).

Taking your analogy of a 50 amp breaker allowing 50 amps of 120 volts on one side and then adding another 50 amps to the other side would be a total of 100 amps of 240 volt series circuit.

I agree with you that a 50 amp feeder will be more than he will ever use but I donâ€™t see any reason not to install 250 kcmil copper conductors and a 250 amp disconnect out there if that is what he wants to do.

After giving what you have said some thought I believe I am beginning to grasp what you are trying to say. I think you are saying one thing and I am interpreting it as something different.

A 2 pole 50 amp breaker will handle 12000 watts of electrical energy at 240 volts therefore either side of this breaker will handle 6000 watts of electrical energy to ground. Or as the way you are putting it 50 amps of current on one side and then 50 amps of opposing current on the other side. This would be 50 amps to ground on one side with 50 amps to ground of opposing current on the other side which would be 50 amps of current at 240 volts.

If this is true then I stand corrected. What I was thinking is 50 plus 50 or a total of 100 amps off of this one 50 amp breaker which cannot happen.

Am I now correct in what you are stating? I have concluded this from your comment earlier (post 28) about the 60-40 ratio on the same 50 amp breaker.

So over the weekend I did my install, I have one wire to each breaker on the house panel box, and one wire to each breaker in the garage panel box. From the house panel box, I have one wire from each of the buss bars, (one neutral, and one ground) to the box in the garage. I did not install the grounding bolt (or whatever the bolt is called I`m sure I`ll get beat up for not knowing its name) so in the garage I have one buss bar for neutral, and one buss bar for ground. On the ground buss bar, I have a bare wire to one grounding rod, and I ran it thru the connector, and have the second grounding rod 10 feet away. In the garage I have one 100 amp main breaker, then there is one 40 amp 220, one 15 amp 110 (Lights), one 15 amp 110 breaker (Wall outlets). I put theses all on the same side of the panel box hoping to balance the load somewhat.
Last night I had 4 fluorescent tube lights (total 8 bulbs) 1 radio, 2 outside 60 watt lights and a box fan. When I fire up my welder and struck an arc, the 40 amp breaker in the house tripped. So I did the obvious, turned off the outside lights, radio, and fan, and was able to weld.
I figured since I used 90 amp wire I would not have very much amperage loss on the 40 amp breaker. Or is it the other way, since I have wire rated for 90 amp, am I loosing amperage from the 40 amp breaker in the main panel?
I purchased 50 feet of cable, and have between 4 and 5 feet cut off, so length of run is not a factor. All breakers are brand new never used.
Is there something I missed in the connection? I have my electrical inspection tomorrow, and want to make sure I have not missed something.

FYI, the only reason i have a 40 amp breaker in the main panel is that the store was out of 50`s, and I want the trench back filled this week. I will be getting a 50 when they get more in.

Next spring when i upgrade the service to my house from 100 amp to 200 amp i am planning to do that......I just needed to get power into the garage for the winter, and wanted to make sure what i was running would accomidate future upgrade.